What are the solutions to the equation (3(x-1)-4x+(x-5))/4 =(-2x(x-2)-4x-7)/2 ?

The solutions to the equation (3(x-1)-4x+(x-5))/4 =(-2x(x-2)-4x-7)/2 are x=sqrt(3/2)i,x=-sqrt(3/2)i

Find the zeros of (3(x-1)-4x+(x-5))/4 =(-2x(x-2)-4x-7)/2

The zeros of (3(x-1)-4x+(x-5))/4 =(-2x(x-2)-4x-7)/2 are x=sqrt(3/2)i,x=-sqrt(3/2)i

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\frac{3 ( x - 1 ) - 4 x + ( x - 5 )}{4} = \frac{- 2 x ( x - 2 ) - 4 x - 7}{2}

x = \frac{3}{2} i, x = - \frac{3}{2} i

Solution steps

\frac{3 ( x - 1 ) - 4 x + ( x - 5 )}{4} = \frac{- 2 x ( x - 2 ) - 4 x - 7}{2}

Expand

\frac{3 ( x - 1 ) - 4 x + ( x - 5 )}{4} : - 2

Expand

\frac{- 2 x ( x - 2 ) - 4 x - 7}{2} : - x^{2} - \frac{7}{2}

- 2 = - x^{2} - \frac{7}{2}

Switch sides

- x^{2} - \frac{7}{2} = - 2

Move

\frac{7}{2}

to the right side

- x^{2} = - 2 + \frac{7}{2}

Simplify

- 2 + \frac{7}{2} : \frac{3}{2}

- x^{2} = \frac{3}{2}

Divide both sides by

- 1

x^{2} = - \frac{3}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = - \frac{3}{2}, x = - - \frac{3}{2}

- \frac{3}{2} = \frac{3}{2} i

- - \frac{3}{2} = - \frac{3}{2} i

x = \frac{3}{2} i, x = - \frac{3}{2} i

9 - ∣ 2 x + 8 ∣ \leq 5

- 6 y - y^{2} = - 28

14 x - 16 = 10 x

y^{2} + 9 = 0

14 = - (p - 8)

What are the solutions to the equation (3(x-1)-4x+(x-5))/4 =(-2x(x-2)-4x-7)/2 ?
The solutions to the equation (3(x-1)-4x+(x-5))/4 =(-2x(x-2)-4x-7)/2 are x=sqrt(3/2)i,x=-sqrt(3/2)i
Find the zeros of (3(x-1)-4x+(x-5))/4 =(-2x(x-2)-4x-7)/2
The zeros of (3(x-1)-4x+(x-5))/4 =(-2x(x-2)-4x-7)/2 are x=sqrt(3/2)i,x=-sqrt(3/2)i